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- Solution first found by St. Venant.
- Tractions at the ends are statically equivalent to equal and opposite torques .
- Lateral surfaces are traction-free.
- An axis passes through the center of twist ( axis).
- Each c.s. projection on to the plane rotates,but remains undistorted.
- The rotation of each c.s. ( ) is proportional to .
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where is the twist per unit length.
- The out-of-plane distortion (warping) is the same for each c.s. and is proportional to .
- Torsional rigidity ( ).
- Maximum shear stress.
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where is the warping function.
If (small strain),
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Therefore,
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Therefore,
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Therefore,
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- Normal to cross sections is .
- Normal traction .
- Projected shear traction is .
- Traction vector at a point in the cross section is tangent to the cross section.
Boundary Conditions on Lateral Surfaces
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- Lateral surface traction-free.
- Unit normal to lateral surface appears as an in-plane unit normal to the boundary .
We parameterize the boundary curve using
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The tangent vector to is
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The tractions and on the lateral surface are identically zero.
However, to satisfy the BC , we need
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or,
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Boundary Conditions on End Surfaces
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The traction distribution is statically equivalent to the torque .
At ,
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Therefore,
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From equilibrium,
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Hence,
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The Green-Riemann Theorem
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If and then
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with the integration direction such that is to the left.
Applying the Green-Riemann theorem to equation (17), and using
equation (16)
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Similarly, we can show that . since .
The moments about the and axes are also zero.
The moment about the axis is
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where is the torsion constant. Since , we have
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If , then , the polar moment of inertia.